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Lemma

Let

$$\mathcal{C} \array{ \overset{ \phantom{A} \Pi \phantom{A} }{\longrightarrow} \\ \overset{ \phantom{A} Disc \phantom{A} }{\hookleftarrow} \\ \overset{ \phantom{A} \Gamma \phantom{A} }{ \longrightarrow } } \mathcal{D}$$

be an adjoint triple, with $Disc$ a fully faithful functor. Denoting the adjunction units/counits as

$\phantom{A}$ adjunction $\phantom{A}$	$\phantom{A}$ unit $\phantom{A}$	$\phantom{A}$ counit $\phantom{A}$
$\phantom{A}$ $(\Pi \dashv Disc)$ $\phantom{A}$	$\phantom{A}$ $\eta^{ʃ}$ $\phantom{A}$	$\phantom{A}$ $\epsilon^{ʃ}$ $\phantom{A}$
$\phantom{A}$ $(Disc \dashv \Gamma)$ $\phantom{A}$	$\phantom{A}$ $\eta^\flat$ $\phantom{A}$	$\phantom{A}$ $\epsilon^\flat$ $\phantom{A}$

then the following composites of unit/counit components are equal:

(1)$$\left( \eta^{\flat}_{\Pi X} \right) \circ \left( \Pi \epsilon^\flat_X \right) \;\;=\;\; \left( \Gamma \eta^{ʃ}_{X} \right) \circ \left( \epsilon^{ʃ}_{\Gamma X} \right) \phantom{AAAAAA} \array{ \Pi Disc \Gamma X &\overset{\epsilon^{ʃ}_{\Gamma X}}{\longrightarrow}& \Gamma X \\ {}^{ \mathllap{ \Pi \epsilon^\flat_X } }\big\downarrow && \big\downarrow^{\mathrlap { \Gamma \eta^{ʃ}_{X} } } \\ \Pi X &\underset{ \eta^\flat_{\Pi X} }{\longrightarrow}& \Gamma Disc \Pi X }$$

Proof

We claim that the following diagram commutes:

$$\array{ && && \Gamma X \\ && & {}^{ \epsilon^ʃ_{\Gamma X} }\nearrow && \searrow^{\mathrlap{ \Gamma \eta^{ʃ}_X }} \\ && \Pi Disc \Gamma X && && \Gamma Disc \Pi X \\ & {}^{ \Pi \epsilon^\flat_X }\swarrow && \searrow^{ \mathrlap{ \Pi Disc \Gamma \eta^{ʃ}_X } } && {}^{\mathllap{ \eta^{ʃ}_{\Gamma Disc \Pi X} }}\nearrow && \nwarrow^{ \mathrlap{ \eta^{\flat}_{\Pi X} } } \\ \Pi X && && \Pi Disc \Gamma Disc \Pi X && && \Pi X \\ & {}_{\mathllap{ \Pi \eta^{ʃ}_X }}\searrow && \swarrow_{\mathrlap{ \Pi \epsilon^{\flat}_{Disc \Pi X} }} && {}_{\mathllap{ \Pi Disc \eta^\flat_{\Pi X} }}\nwarrow && \nearrow_{\mathrlap{ \epsilon^{ʃ}_{\Pi X} }} \\ && \Pi Disc \Pi X && \underset{id_{\Pi Disc \Pi X}}{\longleftarrow} && \Pi Disc \Pi X }$$

This commutes, because:

1. the left square is the image under $\Pi$ of naturality for $\epsilon^\flat$ on $\eta^{ʃ}_X$;

2. the top square is naturality for $\epsilon^{ʃ}$ on $\Gamma \eta^{ʃ}_X$;

3. the right square is naturality for $\epsilon^{ʃ}$ on $\eta^{\flat}_{\Pi X}$;

4. the bottom commuting triangle is the image under $\Pi$ of the zig-zag identity for $(Disc \dashv \Gamma)$ on $\Pi X$.

Finally, also the total bottom composite is the identity morphism $id_{\Pi X}$, due to the zig-zag identity for $(ʃ \dashv Disc)$.

Therefore the total composite from $\Pi Disc \Gamma X \to \Gamma Disc \Pi X$ along the bottom part of the diagram equals the left hand side of (1), while the composite along the top part of the diagram clearly equals the right hand side of (1).

Proposition

(points-to-pieces transform)

Consider an adjoint quadruple of the form

$$\Pi \dashv Disc \dashv \Gamma \dashv coDisc \;\;\colon\;\; \mathbf{H} \array{ \overset{\phantom{AA} \Pi \phantom{AA} }{\longrightarrow} \\ \overset{\phantom{AA} Disc \phantom{AA} }{\hookleftarrow} \\ \overset{\phantom{AAA} \Gamma \phantom{AAA} }{\longrightarrow} \\ \overset{\phantom{AA} coDisc \phantom{AA} }{\hookleftarrow} } \mathbf{B}$$

(for instance a cohesive topos over some base topos $\mathbf{B}$).

Then for all $X \in \mathbf{X}$ the following two natural transformations, constructed from the adjunction units/counits and their inverse morphisms (using idempotency), are equal:

(2)$$ptp_{\mathbf{B}} \;\;\coloneqq\;\; \left( \Pi \epsilon^\flat_X \right) \circ \left( \eta^{ʃ}_{\Gamma X} \right)^{-1} \;\;=\;\; \left( \eta^\flat_{\Pi X} \right)^{-1} \circ \left( \Gamma \eta^{ʃ}_X \right) \phantom{AAAAAAA} \array{ \Gamma X & \overset{ \Gamma \eta^{ʃ}_X }{\longrightarrow} & \Gamma Disc \Pi X \\ {}^{ \mathllap{ \left( \eta^{ʃ}_{\Gamma X} \right)^{-1} } }\big\downarrow & \searrow^{ \mathrlap{ ptp_{\mathbf{B}} } } & \big\downarrow^{ \mathrlap{ \left( \eta^\flat_{\Pi X} \right)^{-1} } } \\ \Pi Disc \Gamma X &\underset{ \Pi \epsilon^\flat_X }{\longrightarrow}& \Pi X }$$

Moreover, the image of these morphisms under $Disc$ equals the following composite:

(3)$$ptp_{\mathbf{H}} \;\colon\; \flat X \overset{ \phantom{A} \epsilon^{\flat}_X \phantom{A} }{\longrightarrow} X \overset{ \phantom{A} \eta^{ʃ}_X \phantom{A} }{\longrightarrow} ʃ X \,,$$

hence

(4)$$ptp_{\mathbf{H}} \;=\; Disc(ptp_{\mathbf{B}}) \,.$$

Either of these morphisms we call the points-to-pieces transform.

Proof

For the second statement, notice that the $(Disc \dashv \Gamma)$-adjunct of

$$ptp_{\mathbf{H}} \;\colon\; Disc \Gamma X \overset{ \phantom{A} \epsilon^{\flat}_X \phantom{A} }{\longrightarrow} X \overset{ \phantom{A} \eta^{ʃ}_X \phantom{A} }{\longrightarrow} Disc \Pi X$$

is

(5)$$\widetilde{ ptp_{\mathbf{H}} } \;\;=\;\; \underset{ = id_{\Gamma X} }{ \underbrace{ \Gamma X \underoverset{iso}{ \phantom{A} \eta^{\flat}_{\Gamma X} \phantom{A} }{ \longrightarrow } \Gamma Disc \Gamma X \underoverset{iso}{ \phantom{A} \Gamma \epsilon^{\flat}_X \phantom{A} }{\longrightarrow} \Gamma X }} \overset{ \phantom{A} \Gamma \eta^{ʃ}_X \phantom{A} }{\longrightarrow} \Gamma Disc \Pi X \,,$$

where under the braces we uses the zig-zag identity.

(As a side remark, for later usage, we observe that the morphisms on the left in (5) are isomorphisms, as shown, by idempotency of the adjunctions.)

From this we obtain the following commuting diagram:

$$\array{ Disc \Gamma X &\overset{ \phantom{A} Disc \Gamma \eta^{ʃ}_X \phantom{A} }{\longrightarrow}& Disc \Gamma Disc \Pi X &\underoverset{iso}{ \phantom{A} Disc \left(\eta^{ \flat }_{\Pi X}\right)^{-1} \phantom{A} }{ \longrightarrow }& Disc \Pi X \\ &{}_{\mathllap{ ptp_{\mathbf{H}} }}\searrow& {}^{ \mathllap{ \epsilon^{\flat}_{Disc \Pi X} } } \big\downarrow^{\mathrlap{\simeq}} & \nearrow_{\mathrlap{id_{\Pi X}}} \\ && Disc \Pi X }$$

Here:

1. on the left we identified $\widetilde {\widetilde {ptp_{\mathbf{H}}}} \;=\; ptp_{\mathbf{\mathbf{H}}}$ by applying the formula for $(Disc \dashv \Gamma)$-adjuncts to $\widetilde {ptp_{\mathbf{H}}} = \Gamma \eta^{ʃ}_X$ (5);

2. on the right we used the zig-zag identity for $(Disc \dashv \Gamma)$.

This proves the second statement.

form the $(Disc \dashv \Gamma)$-adjoint:

$$\underset{ = id_{\Gamma X} }{ \underbrace{ \Gamma X \underoverset{iso}{ \phantom{A} \eta^{\flat}_{\Gamma X} \phantom{A} }{ \longrightarrow } \Gamma Disc \Gamma X \underoverset{iso}{ \phantom{A} \Gamma \epsilon^{\flat}_X \phantom{A} }{\longrightarrow} \Gamma X }} \overset{ \phantom{A} \Gamma \eta^{ʃ}_X \phantom{A} }{\longrightarrow} \Gamma Disc \Pi X$$

and postcompose

$$\Gamma X \overset{ \phantom{A} \Gamma \eta^{ʃ}_X \phantom{A} }{\longrightarrow} \Gamma Disc \Pi X \underoverset{iso}{ \phantom{A} \left(\eta^{ \flat }_{\Pi X}\right)^{-1} \phantom{A} }{ \longrightarrow } \Pi X$$

alternatively, form the $(\Pi \dashv Disc)$-adjoint

$$\Pi Disc \Gamma X \overset{ \phantom{A} \Pi \epsilon^{\flat}_X \phantom{A} }{\longrightarrow} \underset{ = id_{\Pi X} }{ \underbrace{ \Pi X \underoverset{iso}{ \phantom{A} \Pi \eta^{ʃ}_X \phantom{A} }{\longrightarrow} \Pi Disc \Pi X \underoverset{iso}{ \phantom{A} \epsilon^{ʃ}_{\Pi X} \phantom{A}}{\longrightarrow} \Pi X } }$$

and precompose

$$\Gamma X \overset{ \phantom{A} \left(\eta^{ʃ}_{\Gamma X}\right)^{-1} \phantom{A} }{\longrightarrow} \Pi Disc \Gamma X \overset{ \phantom{A} \Pi \epsilon^{\flat}_X \phantom{A} }{\longrightarrow} \Pi X$$

and more

$$\array{ \Pi Disc \Gamma X &\overset{ \phantom{A} \Pi \epsilon^{\flat}_X \phantom{A} }{\longrightarrow} & \Pi X \\ {}^{ \mathllap{ \Pi Disc \Gamma \eta^\flat_X } } \big\downarrow^{ \mathrlap{\simeq} } && {}^{\mathllap{\simeq}}\big\downarrow^{\mathrlap{ \Pi \eta^\flat_X }} \\ \Pi Disc \Gamma Disc \Gamma X & \underset{ \Pi \epsilon^{\flat}_{Disc \Gamma X} }{\longrightarrow}& \Pi Disc \Gamma X }$$

now consider

$$Disc X \overset{ \phantom{A} \eta^{\sharp}_{Disc X} \phantom{A} }{\longrightarrow} \sharp Disc X$$

hence

$$Disc X \overset{ \phantom{A} \eta^{\sharp}_{Disc X} \phantom{A} }{\longrightarrow} coDisc \Gamma Disc X$$

and postcompose

$$Disc X \overset{ \phantom{A} \eta^{\sharp}_{Disc X} \phantom{A} }{\longrightarrow} coDisc \Gamma Disc X \overset{ coDisc \left( \eta^{\flat}_{X} \right)^{-1} }{ \longrightarrow} coDisc X$$

Lemma

Consider an adjoint triple

$$Disc \dashv \Gamma \dashv coDisc \;\;\colon\;\; \mathbf{H} \array{ \overset{\phantom{AA} Disc \phantom{AA} }{\longleftarrow} \\ \overset{\phantom{AAA} \Gamma \phantom{AAA} }{\longrightarrow} \\ \overset{\phantom{AA} coDisc \phantom{AA} }{\longleftarrow} } \mathbf{B}$$

Then application of the functor $\Gamma$ on any morphism $\mathbf{X} \overset{f}{\to} \mathbf{Y} \;\;\in \mathbf{H}$ is equal to the operations of

1. pre-composition with the $(Disc \dashv \Gamma)$-adjunction counit $\epsilon^\flat_{\mathbf{X}}$, followed by passing to the $(Disc \dashv \Gamma)$-adjunct;

2. post-composition with the $(\Gamma \dashv coDisc)$-adjunction unit $\eta^{ \sharp }_{\mathbf{Y}}$, followed by passing to the $(\Gamma \dashv coDisc)$-adjunct:

(6)$$\Gamma_{\mathbf{X}, \mathbf{Y}} \;=\; \widetilde{\eta^\sharp_{\mathbf{Y}} \circ (-)} \;=\; \widetilde{ (-) \circ \epsilon^\flat_{\mathbf{X}} } \,.$$

Proof

For the first equality, consider the following naturality square for the adjunction hom-isomorphism (this Def.):

$$\array{ Hom_{\mathbf{B}}( \Gamma \mathbf{Y} , \Gamma \mathbf{Y} ) &\overset{\widetilde {(-)}}{\longrightarrow}& Hom_{\mathbf{H}}( \mathbf{Y}, coDisc \Gamma \mathbf{Y} ) \\ {}^{\mathllap{ Hom_{\mathbf{B}}(\Gamma(f), \Gamma \mathbf{Y}) }} \big\downarrow && \!\!\!\!\! \big\downarrow^{\mathrlap{ Hom_{\mathbf{H}}( f, coDisc \Gamma \mathbf{Y} ) }} \\ Hom_{\mathbf{B}}( \Gamma \mathbf{X}, \Gamma \mathbf{Y} ) &\overset{\widetilde{ (-) }}{\longleftarrow}& Hom_{\mathbf{H}}( \mathbf{X}, coDisc \Gamma \mathbf{Y} ) } \phantom{AAAAA} \array{ \{ \Gamma \mathbf{Y} \overset{id_{\Gamma \mathbf{Y}}}{\to} \Gamma \mathbf{Y}\} &\longrightarrow& \{ \mathbf{Y} \overset{\eta^\sharp_{\mathbf{Y}}}{\to} coDisc \Gamma \mathbf{Y} \} \\ \big\downarrow && \big\downarrow \\ \{ \Gamma \mathbf{X} \overset{\Gamma(f)}{\to} \Gamma \mathbf{Y} \} &\longleftarrow& \{ \mathbf{X} \overset{\eta^\sharp_{\mathbf{Y}} \circ f}{\longrightarrow} coDisc \Gamma \mathbf{Y} \} }$$

Chasing the identity morphism $id_{\Gamma \mathbf{Y}}$ through this diagram, yields the claimed equality, as shown on the right. Here we use that the right adjunct of the identity morphism is the adjunction unit, as shown.

The second equality is fomally dual:

$$\array{ Hom_{\mathbf{B}}( \Gamma \mathbf{X}, \Gamma \mathbf{X}) &\overset{\widetilde { (-) }}{\longrightarrow}& Hom_{\mathbf{H}}( Disc \Gamma \mathbf{X} , \mathbf{X}) \\ {}^{\mathllap{ Hom_{\mathbf{B}}( \Gamma \mathbf{X}, \Gamma(f) ) }} \big\downarrow && \big\downarrow^{ \mathrlap{ Hom_{\mathbf{X}}( Disc \Gamma \mathbf{X}, f ) } } \\ Hom_{\mathbf{B}}( \Gamma \mathbf{X}, \Gamma \mathbf{Y} ) &\overset{ \widetilde{ (-) } }{\longleftarrow}& Hom_{\mathbf{H}}( Disc \Gamma \mathbf{X}, \mathbf{Y} ) } \phantom{AAAAA} \array{ \{ \Gamma \mathbf{X} \overset{id_{\Gamma \mathbf{X}}}{\to} \Gamma \mathbf{X} \} &\longrightarrow& \{ Disc \Gamma \mathbf{X} \overset{\epsilon^{\flat}_X}{\to} \mathbf{X} \} \\ \big\downarrow && \big\downarrow \\ \{ \Gamma \mathbf{X} \overset{\Gamma(f)}{\to} \Gamma(\mathbf{Y}) \} &\longleftarrow& \{ Disc \Gamma \mathbf{X} \overset{f\circ \epsilon^\flat_{\mathbf{X}} }{\longrightarrow} \mathbf{Y}\} }$$

Proposition

Consider an adjoint quadruple of the form

$$\Pi \dashv Disc \dashv \Gamma \dashv coDisc \;\;\colon\;\; \mathbf{H} \array{ \overset{\phantom{AA} \Pi \phantom{AA} }{\longrightarrow} \\ \overset{\phantom{AA} Disc \phantom{AA} }{\hookleftarrow} \\ \overset{\phantom{AAA} \Gamma \phantom{AAA} }{\longrightarrow} \\ \overset{\phantom{AA} coDisc \phantom{AA} }{\hookleftarrow} } \mathbf{B}$$

(for instance a cohesive topos over some base topos $\mathbf{B}$).

Then the following are equivalent:

1. pieces have points:

   1. as seen in $\mathbf{B}$: For every object $X \in \mathbf{X}$ the points-to-pieces transform (Prop. 1) in $\mathbf{B}$ (2) is an epimorphism:
   $$ptp_{\mathbf{B}} \;\colon\; \Gamma X \overset{ epi }{\longrightarrow} \Pi X$$
   1. equivalently, as seen in $\mathbf{H}$: For every object $X \in \mathbf{X}$ the points-to-pieces transform (Prop. 1) in $\mathbf{H}$ (3) is an epimorphism:
   $$ptp_{\mathbf{H}} \;\colon\; \flat X \overset{ epi }{\longrightarrow} ʃ X$$

2. discrete objects are concrete: For every object $S \in \mathbf{B}$ the discrete object $Disc(S)$ is a concrete object, in that the sharp adjunction counit on $Disc(S)$ is a monomorphism:

   $$\eta^\sharp_{Disc(S)} \;\colon\; Disc S \overset{ mono }{\longrightarrow} \sharp Disc S$$

Proof

First observe the equivalence of the first two statements:

$$ptp_{\mathbf{H}} \;\; \text{is epi} \phantom{AAA} \text{iff} \phantom{AAA} ptp_{\mathbf{B}} \;\; \test{is epi} \,.$$

In one direction, assume that $ptp_{\mathbf{B}}$ is an epimorphism. By (4) we have $ptp_{\mathbf{H}} = Disc(ptp_{\mathbf{B}})$, but $Disc$ is a left adjoint and left adjoints preserve monomorphisms.

In the other direction, assume that $ptp_{\mathbf{H}}$ is an epimorphism. By (2) and (5) we see that $ptp_{\mathbf{B}}$ is re-obtained from this by applying $\Gamma$ and then composition with isomorphisms. But $\Gamma$ is again a left adjoint, and hence preserves epimorphism, as does composition with isomorphisms.

By applying (2) again, we find in particular that pieces have points is also equivalent to $\Pi \epsilon^\flat_{Disc S}$ being an epimorphism, for all $S \in \mathbf{B}$. But this is equivalent to

$$Hom_{\mathbf{B}}(\Pi \epsilon^\flat_{\mathbf{X}}, S) = Hom_{\mathbf{\mathbf{H}}}(\epsilon^\flat_{\mathbf{X}}, Disc(S))$$

being a monomorphism for all $S$ (by adjunction isomorphism and definition of epimorphism).

Now by Lemma 2, this is equivalent to

$$Hom_{\mathbf{H}}( \mathbf{X}, \eta^\sharp_{Disc(S)} )$$

being a monomorphism, which is equivalent to $\eta^\sharp_{Disc(S)}$ being a monomorphism, hence to discrete objects are concrete.